Help needed please!

**doctorgk** · April 26th 2008, 02:52 PM

A rectangle made of elastic material will be made into a cylinder by joining edges AD and BC. To support the structure, a wire of fixed length L is placed along the diagonal of the rectangle. Find the angle $\theta$ that will result in the cylinder of maximum volume.

see attached picture

thank you very much!

unfortunately i cannot attach a picture
so here is a shot at it...

d._________c
..|............/|
..|........./...|
..|....../......|
..|.../.........|
a|/............|b
theta is angle a, in the triangle cab

**Soroban** · April 27th 2008, 09:06 AM

Hello, doctorgk!

A rectangle made of elastic material will be made into a cylinder by joining edges AD and BC.
To support the structure, a wire of fixed length L is placed along the diagonal of the rectangle.
Find the angle $\theta\;(\angle CAB)$ that will maximize the volume of the cylinder.

Code:

    D * - - - - - - - * C
      |             * |
      |           *   |
      |      L  *     |
      |       *       | L·sinθ
      |     *         |
      |   *           |
      | * θ           |
    A * - - - - - - - * B
           L·cosθ

We have: . $AB \,=\,L\cos\theta,\;CB \,=\,L\sin\theta$

The height of the cylinder is: . $h \:=\:CB \:=\:L\sin\theta\;\;{\color{blue}[1]}$

$AB$ is the circumference of the circular base of the cylinder.
. . $L\cos\theta \,=\,2\pi r\quad\Rightarrow\quad r \:=\:\frac{L\cos\theta}{2\pi}\;\;{\color{blue}[2]}$

The volume of a cylinder is: . $V \;=\;\pi r^2h\;\;{\color{blue}[3]}$

Substitute [1] and [2] into [3]: . $V \;=\;\pi\left(\frac{L\cos\theta}{2\pi}\right)^2(L\ sin\theta)$

. . and we have: . $\boxed{V \;=\;\frac{L^3}{4\pi}\,\sin\theta\cos^2\!\theta}$

And that is the function to be maximized . . . good luck!

**doctorgk** · April 27th 2008, 11:33 AM

i got it so far, but just a quick question on the optimization. when u differentiate would you factor out the (L^3)/4 pi and then differentiate the trig functions, or would you differentiate the entire thing?

**Moo** · April 27th 2008, 12:40 PM

Hello,

I think it's better factorising, less obstruction, more visibility

**doctorgk** · April 27th 2008, 01:37 PM

so i got the derivative to be...
$\frac{dV}{d\theta}=(\frac{L^3}{4\pi})\cos^3\theta - \sin\theta2\cos\theta\sin\theta$

then i set it equal to zero and i get

$0=\cos^3\theta - \sin\theta2\cos\theta\sin\theta$

what do i do from here?

**Moo** · April 27th 2008, 01:43 PM

Originally Posted by doctorgk

so i got the derivative to be...
$\frac{dV}{d\theta}=(\frac{L^3}{4\pi}) {\color{red}(}\cos^3\theta + \sin\theta2\cos\theta\-sin\theta{\color{red})}$

then i set it equal to zero and i get

$0=\cos^3\theta + \sin\theta2\cos\theta$

what do i do from here?

Hm... No... You have to keep the factor L^3/4pi for the whole thing (correction in red)

The derivative will be :
$\underbrace{\cos^3(\theta)}_{\text{You found it}}+\sin(\theta) 2 (\underbrace{-\sin(\theta)}_{\text{derivative of cos}}) \cos(\theta)$

**doctorgk** · April 27th 2008, 01:51 PM

so then the derivative is

$0=\cos^3\theta - \sin\theta2\cos\theta\sin\theta$

then i can factor out a cosine and get
$0=\cos^2\theta - 2\sin^2\theta$

i know u have to solve for theta, but i kinda forgot how to solve trig equations